PTA Deque (15 分)

3-3 Deque (15 分)

A “deque” is a data structure consisting of a list of items, on which the following operations are possible:

Push(X,D): Insert item X on the front end of deque D.
Pop(D): Remove the front item from deque D and return it.
Inject(X,D): Insert item X on the rear end of deque D.
Eject(D): Remove the rear item from deque D and return it. Write routines to support the deque that take O(1) time per operation.

Format of functions:

Deque CreateDeque();
int Push( ElementType X, Deque D );
ElementType Pop( Deque D );
int Inject( ElementType X, Deque D );
ElementType Eject( Deque D );

where Deque is defined as the following:

typedef struct Node *PtrToNode;
struct Node {
    ElementType Element;
    PtrToNode Next, Last;
};
typedef struct DequeRecord *Deque;
struct DequeRecord {
    PtrToNode Front, Rear;
};

Here the deque is implemented by a doubly linked list with a header. Front and Rear point to the two ends of the deque respectively. Front always points to the header. The deque is empty when Front and Rear both point to the same dummy header. Note: Push and Inject are supposed to return 1 if the operations can be done successfully, or 0 if fail. If the deque is empty, Pop and Eject must return ERROR which is defined by the judge program.

Sample program of judge:

#include <stdio.h>
#include <stdlib.h>

#define ElementType int
#define ERROR 1e5
typedef enum { push, pop, inject, eject, end } Operation;

typedef struct Node *PtrToNode;
struct Node {
    ElementType Element;
    PtrToNode Next, Last;
};
typedef struct DequeRecord *Deque;
struct DequeRecord {
    PtrToNode Front, Rear;
};
Deque CreateDeque();
int Push( ElementType X, Deque D );
ElementType Pop( Deque D );
int Inject( ElementType X, Deque D );
ElementType Eject( Deque D );

Operation GetOp();          /* details omitted */
void PrintDeque( Deque D ); /* details omitted */

int main()
{
    ElementType X;
    Deque D;
    int done = 0;

    D = CreateDeque();
    while (!done) {
        switch(GetOp()) {
        case push: 
            scanf("%d", &X);
            if (!Push(X, D)) printf("Memory is Full!\n");
            break;
        case pop:
            X = Pop(D);
            if ( X==ERROR ) printf("Deque is Empty!\n");
            break;
        case inject: 
            scanf("%d", &X);
            if (!Inject(X, D)) printf("Memory is Full!\n");
            break;
        case eject:
            X = Eject(D);
            if ( X==ERROR ) printf("Deque is Empty!\n");
            break;
        case end:
            PrintDeque(D);
            done = 1;
            break;
        }
    }
    return 0;
}

/* Your function will be put here */

Sample Input:

Pop
Inject 1
Pop
Eject
Push 1
Push 2
Eject
Inject 3
End
结尾无空行

Sample Output:

Deque is Empty!
Deque is Empty!
Inside Deque: 2 3
结尾无空行

分析：题目想让我们用一个链表实现头部添加，头部删除，尾部添加，尾部删除四种方法，且时间复杂度为O(1)。判断为空的方式为：头指针和尾指针指向同一个结点。双链表嘛。一个结点两个指针,一个指向后一个元素,一个指向前一个元素.
AC代码如下:

Deque CreateDeque(){
	Deque de = malloc(sizeof(struct DequeRecord));
	de->Rear = malloc(sizeof(struct Node));
	de->Front = de->Rear;
	de->Front->Next = NULL;
	de->Front->Last = NULL;
	return de;
}
//压栈
int Push( ElementType X, Deque D ){
	PtrToNode temp = malloc(sizeof(struct Node));
	if(!temp)
		return 0;
	temp->Element = X;
	/*空栈时*/
	if( D->Front == D->Rear){
		temp->Next = NULL;
		temp->Last=D->Front;
		D->Front->Next = temp;
		D->Rear = temp;
	}
	else{
		
		temp->Last = D->Front;
		temp->Next = D->Front->Next;
		D->Front->Next->Last = temp;
		D->Front->Next = temp;
	}
	return 1;
}
//弹栈
ElementType Pop( Deque D ){
	//空栈
	if(D->Front == D->Rear){
		return ERROR;
	}
	int reX = D->Front->Next->Element;
	D->Front->Next = D->Front->Next->Next;
	if(!D->Front->Next){
		D->Rear = D->Front;
	}else{
		D->Front->Next->Last = D->Front;
	}
	return reX;
}
int Inject( ElementType X, Deque D ){
	PtrToNode temp = malloc(sizeof(struct Node));
	if(!temp)
		return 0;
	temp->Element = X;
	temp->Next = NULL;
	temp->Last = D->Rear;
	D->Rear->Next = temp;
	D->Rear = temp;
	return 1;
	
}
ElementType Eject( Deque D ){
	if(D->Front == D->Rear)
	return ERROR;
	int reX = D->Rear->Element;
	D->Rear = D->Rear->Last;
	D->Rear->Next = NULL;
    return reX;
}